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Câu hỏi: Cho hàm số ${f(x)}$ có đạo hàm ${f{\prime}(x)=\cos x+1,\forall x\in\mathbb{R}}$ và ${\displaystyle\int\limits_0^{\dfrac{\pi}{2}}f(x)\mathrm{ d}x=\dfrac{\pi^2}{8}+1}$, khi đó ${f\left(\dfrac{\pi}{2}\right)}$ bằng
A. ${\dfrac{\pi}{2}}$.
B. ${\dfrac{\pi}{2}+1}$.
C. ${\dfrac{\pi}{2}-1}$.
D. ${1}$.
A. ${\dfrac{\pi}{2}}$.
B. ${\dfrac{\pi}{2}+1}$.
C. ${\dfrac{\pi}{2}-1}$.
D. ${1}$.
Ta có $f\left( x \right)=\int{\left( \cos x+1 \right)dx}=\sin x+x+C$.
Thay vào ta được $\int\limits_{0}^{\dfrac{\pi }{2}}{f}(x)\text{d}x=\dfrac{{{\pi }^{2}}}{8}+1\Leftrightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\left( \sin x+x+C \right)}dx=\dfrac{{{\pi }^{2}}}{8}+1$
$\Leftrightarrow \left. \left( -\cos x+\dfrac{{{x}^{2}}}{2}+Cx \right) \right|_{0}^{\dfrac{\pi }{2}}=\dfrac{{{\pi }^{2}}}{8}+1\Leftrightarrow 1+\dfrac{{{\pi }^{2}}}{8}+C.\dfrac{\pi }{2}=\dfrac{{{\pi }^{2}}}{8}+1\Leftrightarrow C=0$.
Vậy $f\left( \dfrac{\pi }{2} \right)=1+\dfrac{\pi }{2}$.
Thay vào ta được $\int\limits_{0}^{\dfrac{\pi }{2}}{f}(x)\text{d}x=\dfrac{{{\pi }^{2}}}{8}+1\Leftrightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\left( \sin x+x+C \right)}dx=\dfrac{{{\pi }^{2}}}{8}+1$
$\Leftrightarrow \left. \left( -\cos x+\dfrac{{{x}^{2}}}{2}+Cx \right) \right|_{0}^{\dfrac{\pi }{2}}=\dfrac{{{\pi }^{2}}}{8}+1\Leftrightarrow 1+\dfrac{{{\pi }^{2}}}{8}+C.\dfrac{\pi }{2}=\dfrac{{{\pi }^{2}}}{8}+1\Leftrightarrow C=0$.
Vậy $f\left( \dfrac{\pi }{2} \right)=1+\dfrac{\pi }{2}$.
Đáp án B.
